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The results are in, and the winners for MMM #8 is. . . Heather Lewis! Thanks to everyone who participated! I wonder which MENSA book she'll choose?

The Answer

Here is Heather's explanation:

The answer is that Martin has a 29.6% chance of making it safely. As an aside, there's a 77.8% chance that at least one route is bully-free but, sadly, much of the time Martin doesn't pick the bully-free route.

I found two ways of solving this problem. One is brute force: There are 4 people and each picks a route (1, 2, or 3) at random, so there are 3^4=81 possible scenarios, all of which are equally likely. I listed those, and then for fun marked which ones had a possible open route (by just checking if the product of the three bullies' routes equaled 6). There were 63 of those. Then I marked in which cases Martin was on a route that none of the bullies took. There were 24 of those. So the probability of there being an open route at all is 63/81 (the 77.8%) and the probability of Martin being safe is 24/81 (the 29.6% chance).

Another way is to use probability. Martin will pick some route home, and the probability for each of Nelson and his brothers (Darryl and Darryl?) picking a different route than Martin is 2/3 [each]. This means there is a (2/3)^3 = 8/27 approx 0.296 chance that Martin gets home safely.

Special Thanks to this week's Sponsor, Mensa!

Before we introduce the problem for this week, we would like to take some time to thank our great sponsor, Mensa.

Mensa is the largest, oldest, and most famous high-IQ society in the world. The non-profit organization restricts its membership to people with high testable IQs. Members must score at the 98th percentile or higher of a standardized, supervised intelligence test.

The people at Mensa have provided Blinkdagger with a bunch of books that range from crossword puzzles, sudoku puzzles, and even a casino gambling guide! We will be handing out these books as the prizes for this contest. For the international participants, we are still offering the 10 dollar gift certificate to Amazon. See the Rules sections for more details! Without further ado . . . let the games begin!

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Rob Slazas, a R&D engineer and 6-sigma blackbelt, joins us again to talk about statistics in MATLAB.

Our first dive into the Statistics Toolbox will cover descriptive statistic. These are the functions that help us understand what data we have on hand. As we said in the introductory post, descriptive statistics stop there, and do not make assumptions or reach conclusions about where your data came from, or how it compares to other datasets.

Contents

Holding Our Breath

So, it turns out that one of Quan's many talents is the ability to hold his breath for long periods of time. I'm wondering if I should challenge him to a breath-holding contest --> whoever holds it longer wins. Since I'm not sure if I have a chance at beating him, I took some practice breath holds and timed myself. Here are my results after 40 holds:

load RobPracticeHolds.mat; y = ones(size(breathholds));
h1 = figure('Position',[100 100 400 100],'Color','w');
scatter(breathholds,y);
% using a scatter plot like a dot plot

xlabel('Seconds');
set(gca,'Yticklabel',[]);
title('Practice Holds');

You can see that my times go from 40-something to over 120 seconds. I'm not very consistent. The middle of the dense cluster of times seems to be around 70 seconds, so let's call that my "expected" time for now. As you can see, viewing these results as a bunch of points has its limitations, so what tools can we employ here to get more precise?

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Hope everyone had a good long weekend. We have a brand new contest rolling over at Sol's blog wildaboutmath.com! His love for infinite series is found in this edition of Monday Math Madness. Sol claims it’s challenging but not brutally difficult, so give it a try. He won’t reveal the source until the contest ends because the answer is posted with the problem.

Thanks to the sponsors for this contest, he has one $25 gift certificate left for the Art of Problem Solving. He also has a couple of Rubik’s Revolutions, courtesy of Techno Source. Depending on how many correct solutions he gets he may give away two prizes!

Wanted Feedback!

Sol, Quan, and I have been enjoying hosting these contests. We love to hear suggestions and comments to make these more entertaining. So let us know if you have something to share.

Our older MATLAB tutorials are consistently being searched on google, and it is a great encouragement to hear people finding them useful. Recently we have our guest blogger Rob Slazas who posted his Intro to Statistics in MATLAB. Please take a look if you haven't and post any requests you may have!

The results are in, and the winners for MMM #6 are . . . Henno Brandsma and Steven King! Hope you guys enjoy your gift certificate to The Art of Problem Solving! The turnout for this contest wasn't nearly as good as the previous one (48 submissions compared to 103), so that tells us that we need to do a bette job advertising and spreading the word. It also tells us that we need a better/more interesting math problem.

For the next contest, we will probably revert back to a single problem. Originally, I thought the idea of having an easier problem and a harder problem would be more appealing to people. Perhaps we'll try to do a double problem again next week! Any feedback would be appreciated!

Once again, we would like to thank our sponsors from The Art of Problem Solving! Stay tuned for the next Monday Math Madness!

Part I Explained

As explained by Efrit Freeq

Each step dilutes the fraction of mayonnaise to 500/510 of what it was at the
previous step. It starts at 100% mayonnaise and we need to dilute it to 50%.

So we need n such that (500/510)^n ~= 1/2

Hence n ~= ln(1/2)/ln(500/510) so n = 35 is as close as we can get.

Part 2 Explained

Explanation by Lieven Marchand

Let us denote by p(n) the number of ways Nortrom can eat n pounds of
beef. Since he will always start by eating either a 1 pound burger or
a two pound burger, p(n) obeys the following recursion:

p(n)=p(n-1)+p(n-2) with initial conditions p(1)=1 and p(2)=2.

These are the well known Fibonacci numbers and so the solution is

p(n)=1/sqrt(5) [((1+sqrt(5))/2)^(n+1) - ((1-sqrt(5))/2)^(n+1)]

This gives us p(17)=2584 and p(25)=121393.